A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in figure.

Q: Why do we only consider one reflection at each plate?

The problem specifically asks for the interference pattern formed by the two beams obtained after *one reflection* at each plate. While multiple internal reflections do occur, their intensities are much lower and are ignored as per the problem's constraints.

Q: Why is the transmission coefficient 75%?

The problem states that each plate reflects 25% of the incident light. Assuming no light is absorbed by the glass, the remaining light must be transmitted. Therefore, $100\% - 25\% = 75\%$ is transmitted.

Q: Does the phase change upon reflection affect the intensity ratio?

A phase change of $\pi$ occurs when light reflects from a denser medium. This phase change determines the positions of the maxima and minima (the fringe pattern) but does not affect the *values* of the maximum and minimum intensities themselves.

Q: Why do we take the square root of the intensities in the formula?

Intensity is proportional to the square of the amplitude ($I \propto A^2$). The principle of superposition applies to amplitudes, not intensities. So, we must convert intensities back to amplitudes ($\sqrt{I}$), add or subtract them, and then square the result to get the final intensity.

Q: What happens to the light that is transmitted through the second plate?

The light transmitted through the second plate continues to travel away from the setup. It does not interfere with the beams reflecting back towards the source, so we don't need to calculate its intensity for this specific question.

Question

A narrow monochromatic beam of light of intensity $I$ is incident on a glass plate as shown in figure. Another identical glass plate is kept close to the first one-and parallel to it. Each glass plate reflects 25 per cent of the light incident on it and transmits the remaining. Find the ratio of the minimum and maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate.

Sigma Percentile · Accepted Answer

### The Dance of Light: Interference from Parallel Glass Plates

Imagine a single beam of light splitting, bouncing, and recombining to create a beautiful pattern of bright and dark fringes. This problem takes us on a journey alongside a light ray as it interacts with two parallel glass plates, testing our understanding of reflection, transmission, and the principle of superposition.

### Tracking the Intensities

When a light beam strikes a surface, it can be reflected, transmitted, or absorbed. In this ideal scenario, we assume no absorption. The problem states that each glass plate reflects $25\%$ of the incident light. This means the reflection coefficient $R$ is $\frac{1}{4}$. Consequently, the remaining $75\%$ of the light is transmitted, giving us a transmission coefficient $T$ of $\frac{3}{4}$.

Let's trace the path of the incident beam, which starts with an initial intensity $I$.

### The Two Interfering Beams

**Ray 1:**
As the incident beam hits the first plate, a portion of it is immediately reflected back. This forms our first interfering beam. Its intensity, $I_1$, is simply $25\%$ of the initial intensity:
$$I_1 = R 	imes I = \frac{I}{4}$$

**Ray 2:**
The remaining $75\%$ of the light transmits through the first plate. The intensity of this transmitted ray is:
$$I_t = T 	imes I = \frac{3I}{4}$$

This transmitted ray travels to the second plate, where it undergoes another reflection. The intensity of the light reflected from the second plate is $25\%$ of $I_t$:
$$I_{r2} = R 	imes I_t = \frac{1}{4} 	imes \frac{3I}{4} = \frac{3I}{16}$$

Finally, this reflected ray travels back and must transmit through the first plate to emerge and interfere with Ray 1. The intensity of this emerging beam, Ray 2, is $75\%$ of $I_{r2}$:
$$I_2 = T 	imes I_{r2} = \frac{3}{4} 	imes \frac{3I}{16} = \frac{9I}{64}$$

### The Master Equation

Now we have the intensities of our two interfering beams: $I_1 = \frac{I}{4}$ and $I_2 = \frac{9I}{64}$.

In an interference pattern, the maximum and minimum intensities depend on the constructive and destructive superposition of the wave amplitudes. Since intensity is proportional to the square of the amplitude ($I \propto A^2$), the amplitudes are proportional to $\sqrt{I}$.

The ratio of the minimum to maximum intensity is given by the square of the difference of their amplitudes divided by the square of their sum:
$$\frac{I_{\min}}{I_{\max}} = \left( \frac{\sqrt{I_1} - \sqrt{I_2}}{\sqrt{I_1} + \sqrt{I_2}} ight)^2$$

### Final Calculation

Let's substitute our calculated intensities into the formula:
$$\frac{I_{\min}}{I_{\max}} = \left( \frac{\sqrt{\frac{I}{4}} - \sqrt{\frac{9I}{64}}}{\sqrt{\frac{I}{4}} + \sqrt{\frac{9I}{64}}} ight)^2$$

Notice how the initial intensity $I$ beautifully cancels out from the numerator and denominator. Taking the square roots, we get:
$$\frac{I_{\min}}{I_{\max}} = \left( \frac{\frac{1}{2} - \frac{3}{8}}{\frac{1}{2} + \frac{3}{8}} ight)^2$$

To subtract and add these fractions, we find a common denominator:
$$\frac{I_{\min}}{I_{\max}} = \left( \frac{\frac{4}{8} - \frac{3}{8}}{\frac{4}{8} + \frac{3}{8}} ight)^2 = \left( \frac{\frac{1}{8}}{\frac{7}{8}} ight)^2$$

Simplifying the fraction, we arrive at our final answer:
$$\frac{I_{\min}}{I_{\max}} = \left( \frac{1}{7} ight)^2 = \frac{1}{49}$$

This elegant result shows how a simple setup of two glass plates can create a highly contrasted interference pattern, a principle widely used in optical instruments like the Fabry-Perot interferometer!

Visualized Solution

$I$

$R and T$

$I_{1}$

$I_{t}$

$I_{r 2}$

$I_{2}$

$\frac{I _{m i n}}{I _{m a x}}$

$Substitution$

$Simplification$

$Final Answer$

$Conclusion$

The Sigma Insight: Interference and Young's Double-Slit Experiment

The Dance of Light

Tracking the Intensities

The Two Interfering Beams

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R and T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } It​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ir2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Imax​Imin​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion